Imágenes de páginas
PDF
EPUB

let

is

Let U, U' be two anautotomic curces of degree n; be the discriminant of the curve U+kU'. Then a polynomial of k of degree 3 (n − 1)'. Let D be the discriminant of with respect to k; then

D= RS3T.

[blocks in formation]

where R is the tact-invariant of U and U'; S=0 is the condition that a determinate number of the curves U + kU' should have a cusp; and T=0 is the condition that such curves should be binodal.

If A is the discriminant of U, the value of is obtained. by writing a+ka', &c., for each coefficient of U in A; hence

where

D = A + kA, + 1⁄2 k2 ▲12+...............(17)

2

▲1 = a'd▲/da + b'd▲ /db + ........... (18),

and since A is of degree 3 (n − 1)'= μ in the coefficients of U, it follows that is an equation of degree μ in k.

(i) Let = 0, then U+kU' has a node, hence :

Through the points of intersection of two anautotomic curves, μ uninodal curves can be drawn.

(ii) Let U be a uninodal curve, then ▲ = 0 and one value of k is zero, which shows that U is one of the curves of the system, hence :

Through the points of intersection of a uninodal and an anautotomic curve, μ-1 other uninodal curves can be drawn.

(iii) Let U be a binodal curve, then by § 6 A=0, and dAda=0, &c. Hence A, 0, and two values of k are zero, accordingly:

=

Through the points of intersection of a binodal and an anautotomic curve μ-2 uninodal curves can be drawn.

The first polar with respect to A of a general anautotomic curve is likewise a general and not a special one; hence the first polar of U with respect to an arbitrary point (f, g, h) is a general curve. In (15) let

U=gU2+hU„, U'=U1, k=ƒ...........(19),

2

then, which is the discriminant of (15), furnishes a relation between (f, g, h) which is the equation of the Steinerian. Also since every coefficient in U is multiplied by g or h, it follows that the degree of every invariant in the coefficients of U is equal to its degree in g and h.

The discriminant D of with respect to f is the equation of the tangents drawn from A to the Steinerian, multiplied by the square of the line joining A to each node, multiplied by the cube of the line joining A to each cusp; hence D is of the required form.

The last two terms of (15) constitute the first polar of U with respect to an arbitrary point 0 on BC; but if 40 is a tangent to the Steinerian, it follows from [§10] that the first polars of A and O must touch one another; hence their tact-invariant must vanish. This gives a relation between g and h which is the equation of the tangents drawn from A to the Steinerian, and we have therefore identified Ras the tact-invariant of U and U'. Since the number of tangents which can be drawn from A to the Steinerian is 3 (n-1) (n-2), this is the degree of R in (g, h) and therefore in the coefficients of U.

Let N be a node on the Steinerian; then the two nodal tangents are the polar lines of two points P and Q, which are the points on the Hessian corresponding to N and are also nodes on the first polar. Accordingly the latter is a binodal curve; and since T is the equation of the line joining A with the nodes on the Steinerian, and the latter possesses 8 = 3 (n − 2) (n − 3) (3n' — 9n −5) nodes, the degree of 7' in g and h is equal to 8, and this gives its degree in the coefficients of U.

If K is a cusp on the Steinerian, the first polar of K has a cusp at the corresponding point on the Hessian; and the first polar is therefore a cuspidal curve. The Steinerian possesses 12 (n-2) (n - 3) cusps, and the degree of S in g and h and also of the coefficients of S is equal to x.

=

In like manner it can be shown that the degrees of R, S and Tin the coefficients of U' are the same as those of U.

The degree of the Steinerian is 3 (n − 2); and therefore the degree of D is 3 (n-2)' (3n-12n+11) in the coefficients. of U and U'; and it is easily verified that

3 (n − 2)* (3n2 - 12n +11) = 3 (n − 1) (n − 2) + 36 (n − 2) (n−3)

[ocr errors]

+3(n-2) (n-3) (3n-9n-5).

Putting n + 1 for n we obtain Salmon's results.

[blocks in formation]
[merged small][merged small][merged small][merged small][merged small][ocr errors]

=

[ocr errors]

dx+ hp, (k). (v. − v) +5, p, (k). (v.' — v ̧') +.....

2!

[ocr errors]

given in my paper on the Euler-Maclaurin Sum Formula* affords a ready method of obtaining approximate values of definite integrals in terms of equidistant values of the function in cases where the differential coefficients, and therefore the differences, rapidly decrease in value.

Taking k=0 and 1, adding the results, and writing h=1, 2, 3, and 6 in succession, we obtain

(«) S1 = [by + Dan + Van + + 30;] = ['v_de+ P+ P+ P.,............

[ocr errors]

+ ¿v,] = ↓ ['v_dx + 2 P ̧+ 23P ̧+23P ̧+.....,

[ocr errors][merged small][merged small][subsumed][ocr errors][ocr errors][subsumed][subsumed][ocr errors]
[ocr errors]

and ba+6n, where n is an integer.

Eliminating P, and P, from (a), (b), and (c), we obtain

5. §, − 48, + § ̧ = '§ ['v.dc + 120P, ̧+..... .

Quarterly Journal of Mathematics, vol. xlvi. (1915), pp. 220–247.

Hence, if fifth and succeeding differential coefficients are

negligible, an approximate value of

-a6n

[ocr errors]

vdx is

[ocr errors]

a

[ocr errors][subsumed]

which is 36P in excess, neglecting terms higher than P.. Weddle's formula is obtained from this by putting n=1. Again, if we eliminate P, and P, from (b), (c), and (d) we obtain

81S,−64S ̧+5S,= 20 ̊ v_dx+25920P ̧ +.........

[merged small][ocr errors]

(2)

[ocr errors]

a

a

v2dx= [.55 (v + 20.8 +20 +13 +...+ Vaton)

a

va

...

+4.05 (Vat2+ Vats +...) — 3.2 (Va+3 + Va 19 +•••+ Va+on-3)], which is 1296 P in excess.

(3)

6

Eliminating P, and P, from S-S,, S,, and S, the result is

a+6n

S v_dx=[.28 (v1+2a+s 8 + 2 va+19+ ··· + Va+on)

a

6

+1.62 (Va+1+Va+3+...) +.58 (Va+3+ Va+9+•••)],

[merged small][merged small][ocr errors][merged small]

v ̧dx = [.28 (v2+Va16) + 1.62 (Va+1 + Va+5)+2. 2va+3],

which is known as Hardy's formula. The error is 64.8P in defect, neglecting terms higher than P.

2. In practical calculations formulæ (1) and (4) are most frequently used, but at the same time it may be useful to have formulæ which give the integral when the interval is most conveniently divided into tenths.

Writing h=1, 2, 5, and 10 in the summation-formula, and taking b=a+10, we have S, and S, as above, and

[ocr errors][merged small][subsumed][subsumed][subsumed][ocr errors][subsumed][subsumed][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small]

From S, S, and S,, we obtain

ra 10

(5) fav da v_dx = 1 1⁄2¢ [8 (va+a+10) + 35 (Va+1 + Va+3+ Va+7 + Va+9)

[blocks in formation]

+125 (Va+1+Va+s+Va+s+ Va+z+ Vaio) +31.],

...

with an error of approximately 3027 P in defect, and from S1, S, and S., we have

[merged small][merged small][ocr errors]

10

[merged small][merged small][ocr errors]

with an error of 10,000 P in excess.

It will be seen that, considering the range, (5) is very little inferior in accuracy to (1) or (4), but (6) and (7) cannot be relied upon to give satisfactory results unless 10,000 P, and succeeding terms are negligible within the limits of accuracy required. Since P-30240 (v-v()) and

=

[blocks in formation]

these formula will all give fairly good results when sixth and higher differences are very small. When this is the case and fourth and fifth differences are also small, the coefficients in (6) and (7) may be modified to give easy formulæ for calculation without seriously affecting the accuracy of the results. The new formulæ are

[blocks in formation]

+ U (Va+1+Ya+3+ Va1s+ Va+7+ Va19) + 38 Va+s]>

with an error of 8P-2036P in excess, and

rat 10

(7a) [** 3

a

[ocr errors]

v ̧dx = [ ? (va+ Va+10) + 1⁄2 (Va+3+ Va+4+ Va+g+ Va+g) − ‡Va+s]

with an error of 20P,+2580 P

... 6

in excess.

« AnteriorContinuar »